Problem: Find $\lim_{x\to 0}\dfrac{\tan(x)}{3x+\tan(x)}$. Choose 1 answer: Choose 1 answer: (Choice A) A $2$ (Choice B) B $\dfrac14$ (Choice C) C $-\dfrac13$ (Choice D) D The limit doesn't exist.
Explanation: Substituting $x=0$ into $\dfrac{\tan(x)}{3x+\tan(x)}$ results in the indeterminate form $\dfrac{0}{0}$. Furthermore, as the expression involves mixed function types, it's not possible to manipulate it algebraically in a way that will help us find the limits. Therefore, we should use L'Hôpital's rule. $\begin{aligned} &\phantom{=}\lim_{x\to 0}\dfrac{\tan(x)}{3x+\tan(x)} \\\\ &=\lim_{x\to 0}\dfrac{\dfrac{d}{dx}[\tan(x)]}{\dfrac{d}{dx}[3x+\tan(x)]} \gray{\text{L'Hôpital's rule}} \\\\ &=\lim_{x\to 0}\dfrac{\sec^2(x)}{3+\sec^2(x)} \\\\ &=\dfrac{\sec^2(0)}{3+\sec^2(0)} \gray{\text{Substitution}} \\\\ &=\dfrac{1}{4} \end{aligned}$ Note that we were only able to use L'Hôpital's rule because the limit $\lim_{x\to 0}\dfrac{\dfrac{d}{dx}[\tan(x)]}{\dfrac{d}{dx}[3x+\tan(x)]}$ actually exists. In conclusion, $\lim_{x\to 0}\dfrac{\tan(x)}{3x+\tan(x)}=\dfrac14$.